Question

An ac source producing emf $\text{e}={{\text{E}}_{0}}\left[ \cos \left( 100\text{ }\\!\\!\pi\\!\\!\text{ t} \right)+\cos \left( 500\text{ }\\!\\!\pi\\!\\!\text{ t} \right) \right]$ is connected in series with a capacitor and a resistor. The steady state current in the circuit is found to be $\text{i}={{\text{I}}_{1}}\cos \left( 100\text{ }\\!\\!\pi\\!\\!\text{ t}+{{\text{ }\\!\\!\theta\\!\\!\text{ }}_{1}} \right)+{{\text{I}}_{2}}\cos \left( 500\text{ }\\!\\!\pi\\!\\!\text{ t}+\text{ }\\!\\!\theta\\!\\!\text{ 2} \right)$
(A) ${{\text{I}}_{1}}>{{\text{I}}_{2}}$
(B) ${{\text{I}}_{1}}\text{=}{{\text{I}}_{2}}$
(C) ${{\text{I}}_{1}}\text{}{{\text{I}}_{2}}$
(D) Nothing can be said

Answer 1

A capacitor is a passive device that stores energy in its electric field and returns energy to the circuit whenever required capacitors do not store Ac voltage
Charge on capacitor is Q=CV
and current $\text{i}=\dfrac{\text{dQ}}{\text{dt}}$

Complete step by step solution
We can assume two different sources connected in series with same orientation
$\begin{aligned} & {{\text{E}}_{1}}={{\text{E}}_{\text{0}}}\cos \text{ 100 }\\!\\!\pi\\!\\!\text{ t} \\\ & {{\text{E}}_{2}}={{\text{E}}_{\text{0}}}\cos \text{ 500 }\\!\\!\pi\\!\\!\text{ t} \\\ \end{aligned}$
Such that
$\begin{aligned} & \text{e}={{\text{E}}_{\text{0}}}\cos \text{ 100 }\\!\\!\pi\\!\\!\text{ t}+{{\text{E}}_{\text{0}}}\cos \text{ 500 }\\!\\!\pi\\!\\!\text{ t} \\\ & \text{ =}{{\text{E}}_{0}}\left[ \cos \text{ 100 }\\!\\!\pi\\!\\!\text{ t}+\cos \text{ 500 }\\!\\!\pi\\!\\!\text{ t} \right] \\\ \end{aligned}$
Becomes,
$\text{e}={{\text{E}}_{1}}+{{\text{E}}_{2}}$
The charge on the capacitor during the steady state is given by
Q=CE
Putting the value of E in this equation
$\text{Q}=\text{C }{{\text{E}}_{0}}\left[ \cos \text{ 100 }\\!\\!\pi\\!\\!\text{ t}+\cos \text{ 500 }\\!\\!\pi\\!\\!\text{ t} \right]$
The steady state current is, thus given by
$\begin{aligned} & \text{i}=\dfrac{\text{dQ}}{\text{dt}} \\\ & \text{i}=\dfrac{\text{d}}{\text{dt}}\left[ \text{C }{{\text{E}}_{0}}\left[ \cos \text{ 100 }\\!\\!\pi\\!\\!\text{ t}+\cos \text{ 500 }\\!\\!\pi\\!\\!\text{ t} \right] \right] \\\ & \text{i}=\dfrac{\text{d}}{\text{dt}}\left( \text{C }{{\text{E}}_{0}}\cos \text{ 100 }\\!\\!\pi\\!\\!\text{ t} \right)+\dfrac{\text{d}}{\text{dt}}\left( \text{C }{{\text{E}}_{0}}\cos \text{ 500 }\\!\\!\pi\\!\\!\text{ t} \right) \\\ & \text{C }{{\text{E}}_{0}}\left( 100\text{ }\\!\\!\pi\\!\\!\text{ } \right)\text{ sin100 }\\!\\!\pi\\!\\!\text{ t}+\text{C }{{\text{E}}_{0}}\left( 500\text{ }\\!\\!\pi\\!\\!\text{ } \right)\text{ sin500 }\\!\\!\pi\\!\\!\text{ t} \\\ & \text{ }\left[ \because \dfrac{\text{d}}{\text{dt}}\cos \text{ t}=\sin \text{ t} \right] \\\ \end{aligned}$
${{\text{i}}_{1}}\text{ sin 100 }\\!\\!\pi\\!\\!\text{ t}+{{\text{i}}_{2}}\text{ sin 500 }\\!\\!\pi\\!\\!\text{ t}$
Where ${{\text{i}}_{1}}=\text{C }{{\text{E}}_{0}}\left( \text{100 }\\!\\!\pi\\!\\!\text{ t} \right)\text{ }$
${{\text{i}}_{2}}=\text{C }{{\text{E}}_{0}}\left( \text{500 }\\!\\!\pi\\!\\!\text{ t} \right)\text{ }$
Now, using the principle of superposition
We get,
$\text{i}={{\text{i}}_{1}}\cos \text{ }\left( \text{100 }\\!\\!\pi\\!\\!\text{ t}+{{\text{ }\\!\\!\theta\\!\\!\text{ }}_{1}} \right)+{{\text{i}}_{2}}\cos \text{ }\left( \text{500 }\\!\\!\pi\\!\\!\text{ t}+{{\text{ }\\!\\!\theta\\!\\!\text{ }}_{2}} \right)\text{ }$
From this expression, we get
$\begin{aligned} & {{\text{i}}_{1}}={{\text{E}}_{0}}\text{100 }\\!\\!\pi\\!\\!\text{ C} \\\ & {{\text{i}}_{2}}={{\text{E}}_{0}}\text{500 }\\!\\!\pi\\!\\!\text{ C} \\\ \end{aligned}$
From this we can easily predict that
${{\text{i}}_{2}}>{{\text{i}}_{1}}$
Therefore option (C) is correct.

Note
Capacitor and battery are different devices because the potential energy in a capacitor is stored in an electric field, where a battery stores its potential energy in a chemical form. Knowledge of some trigonometric rules and differentiation rule should be must be careful while solving the differentiation because students get confused easily